The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. Practice Paper 1 Question 16

Dividing $$x$$ by a small annual $$r$$-percent cumulative interest rate approximates the number of years needed to double your investment with a bank. Find $$x$$. Hint: The word "small" may be important.

The above links are provided as is. They are not affiliated with the Climb Foundation unless otherwise specified.

Warm-up Questions

1. For which real values of $$a$$ does $$(1+a)^x=e$$ yield real values of $$x$$?
2. If the number of bacteria in a sample doubles every hour, how many are there after 10 hours, if the initial population was 1?
3. Using Taylor expansion about 0 (Maclaurin), find $$\sin(0.1)$$ correct to 2 decimal places.

Hints

• Hint 1
For $$r$$-percent cumulative interest every year, how much money will you have after $$n$$ years?
• Hint 2
If the amount of money doubles after $$n$$ years, find an expression relating $$n$$ to $$r$$.
• Hint 3
Can you use the fact that $$r$$ is small and the Taylor series for $$\ln(1+x)$$ to simplify your expression?

Solution

[Trivia: Luca Pacioli in 1494 said "72" (for slightly larger $$r$$), and this result was apparently already well known at that time.]

The question asks us to find $$x$$, which when divided by $$r$$ approximates the number of years needed to double the investment. This translates to $$\frac{x}{r} = n.$$ Let the initial investment amount be $$a_0$$, and the amount in the bank after $$n$$ years be $$a_n$$. Each year, the amount increases by $$r$$ percent, which gives us the recurrence $$a_{n+1} = a_n\big(1+\frac{r}{100}\big).$$ Unwinding this recursion, one finds that $$a_n = a_0 \big(1+\frac{r}{100}\big)^n.$$

We wish to find $$n$$ such that $$a_n = 2a_0.$$ Substitute above to obtain $$2 = \big(1+\frac{r}{100}\big)^n,$$ and apply log to get $$n = \frac{\ln 2}{\ln(1+\frac{r}{100})}.$$ Since $$r$$ is very small, we can approximate the denominator by using only the first term of the Taylor expansion of $$\ln(1+\frac{r}{100}) \approx \frac{r}{100},$$ which gives us $$n = \frac{100 \ln 2} {r}.$$ We now substitute into the previous equation to get $$x = 100 \ln 2 \approx 69.3.$$

The Taylor expansion of $$\ln(1+x)$$ around $$x=0$$ is $$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$$.

If you have queries or suggestions about the content on this page or the CSAT Practice Platform then you can write to us at . Please do not write to this address regarding general admissions or course queries.